POJ1833 & POJ3187 & POJ3785 next_permutation应用

要是没有next_permutation这个函数，这些题认为还不算特别水，只是也不一定，那样可能就会有对应的模板了。

反正正是由于next_permutation这个函数。这些题包含之前的POJ1226，都变得简单起来。

Description

Input

Output

Sample Input

3
3 1
2 3 1
3 1
3 2 1
10 2	
1 2 3 4 5 6 7 8 9 10

Sample Output

3 1 2
1 2 3
1 2 3 4 5 6 7 9 8 10

直接用next_permutation这个函数就可以。

代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;

int num[1030];

int main()
{
	int Test,N,Q,i;
	cin>>Test;
	while(Test--)
	{
		scanf_s("%d%d",&N,&Q);
		for(i=0;i<N;i++)
			scanf_s("%d",&num[i]);
		for(i=1;i<=Q;i++)
			 next_permutation(num,num+N);
		for(i=0;i<N;i++)
			printf("%d ",num[i]);
		printf("\n");
	}
	return 0;
}

Description

    3   1   2   4

      4   3   6

        7   9

         16

Input

Output

Sample Input

4 16

Sample Output

3 1 2 4

Hint

题意是要输出N个数，这N个数是从1到N这些数的一个顺序。这种顺序依照杨辉三角的模式相加起来等于sum，输出相等时的第一个字典顺序。

一看到N是大于1小于10的我就想暴力了。。

。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
	int i,n,sum,a[12];
	cin>>n>>sum;

	for(i=1;i<=10;i++)
		a[i]=i;
	if(n==1)
	{
		cout<<1<<endl;
	}
	else if(n==2)
	{
		cout<<1<<" "<<2<<endl;
	}
	else if(n==3)
	{
		while(1*a[1]+2*a[2]+1*a[3]!=sum)
		{
			next_permutation(a+1,a+3+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<endl;
	}
	else if(n==4)
	{
		while(1*a[1]+3*a[2]+3*a[3]+1*a[4]!=sum)
		{
			next_permutation(a+1,a+4+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<endl;
	}
	else if(n==5)
	{
		while(1*a[1]+4*a[2]+6*a[3]+4*a[4]+1*a[5]!=sum)
		{
			next_permutation(a+1,a+5+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<endl;
	}
	else if(n==6)
	{
		while(1*a[1]+5*a[2]+10*a[3]+10*a[4]+5*a[5]+1*a[6]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<endl;
	}
	else if(n==7)
	{
		while(1*a[1]+6*a[2]+15*a[3]+20*a[4]+15*a[5]+6*a[6]+1*a[7]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<endl;
	}
	else if(n==8)
	{
		while(1*a[1]+7*a[2]+21*a[3]+35*a[4]+35*a[5]+21*a[6]+7*a[7]+1*a[8]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<endl;
	}
	else if(n==9)
	{
		while(1*a[1]+8*a[2]+28*a[3]+56*a[4]+70*a[5]+56*a[6]+28*a[7]+8*a[8]+1*a[9]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<endl;
	}
	else if(n==10)
	{
		while(1*a[1]+9*a[2]+36*a[3]+84*a[4]+126*a[5]+126*a[6]+84*a[7]+36*a[8]+9*a[9]+1*a[10]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<" "<<a[10]<<endl;
	}

	return 0;
}

Description

Input

Output

Sample Input

3 
1 123 
2 279134399742 
3 987

Sample Output

1 132
2 279134423799
3 BIGGEST

还是直接使用next_permutation。这个函数是有返回值的，返回值是0时表示已经没有下一个字典顺序了，它要变成第一个字典顺序。返回值是1时表示还有字典顺序的下一个顺序，所以利用函数的这个性质就OK了。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
	int Test,num;
	char s[100];

	cin>>Test;
	while(Test--)
	{
		cin>>num>>s;
		cout<<num<<" ";

		int n=next_permutation(s,s+strlen(s));
		
		if(n==0)
			cout<<"BIGGEST"<<endl;
		else
			cout<<s<<endl;
	}
	return 0;
}